Question

Find the acute angle θ such that 5tan² θ + 3 = 9sec θ.

Answer 1

5tan θ + 3 = 9sec θ 

∴ 5(sec² θ – 1) + 3 = 9sec θ 

∴ 5sec² θ – 5 + 3 = 9sec θ 

∴ 5sec² θ – 9sec θ – 2 = 0 

∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0 

∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0 

∴ (sec θ – 2) (5sec θ + 1) = 0 

∴ sec θ – 2 = 0 or 5sec θ + 1 = 0 

∴ sec θ = 2 or sec θ = -1/5 

Since sec θ ≥ 1 or sec θ ≤ -1, 

sec θ = 2 

∴ θ = 60° … [ ∵ sec 60° = 2]